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Chapter 1: Problem 13

The wavelength range for infrared radiation is \(10^{-5} \mathrm{m},\) while that of ultraviolet radiation is \(10^{-8} \mathrm{m}\) . Which type of radiation has more energy, and why? (A) Ultraviolet has more energy because it has a higher frequency. (B) Ultraviolet has more energy because it has a longer wavelength. (C) Infrared has more energy because it has a lower frequency. (D) Infrared has more energy because it has a shorter wavelength.

Short Answer

Expert verified
Ultraviolet has more energy because it has a shorter wavelength. Hence, option (A) is the correct answer.

Step by step solution

01

Understand wave properties

Know that the energy of electromagnetic radiation is inversely proportional to the wavelength, which means shorter wavelengths correspond to higher energy.
02

Compare wavelengths

Compare the given wavelengths of ultraviolet and infrared radiation. The wavelength of ultraviolet radiation is \(10^{-8} \mathrm{m}\) which is shorter than the wavelength of infrared radiation, \(10^{-5} \mathrm{m}\).
03

Conclude based on energy-wavelength relationship

Based on the relationship of wavelength and energy, we conclude that the type of radiation with the shorter wavelength has more energy. Thus, ultraviolet radiation has more energy than infrared radiation.

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Most popular questions from this chapter

Use the following information to answer questions 25-28. A voltaic cell is created using the following half-cells: \(\begin{array}{ll}{\mathrm{Cr}^{3+}+3 e \rightarrow \mathrm{Cr}(s)} & {E^{\circ}=-0.41 \mathrm{V}} \\ {\mathrm{Pb}^{2+}+2 e \rightarrow \mathrm{Pb}(s)} & {E^{\circ}=-0.12 \mathrm{V}}\end{array}\) The concentrations of the solutions in each half-cell are 1.0 M. Which of the following best describes the activity in the salt bridge as the reaction progresses? (A) Electrons flow through the salt bridge from the \(\mathrm{Pb} / \mathrm{Pb}^{2+}\) half-cell to the \(\mathrm{Cr} / \mathrm{Cr}^{3+}\) half-cell. (B) \(\quad \mathrm{Pb}^{2+}\) flows to the \(\mathrm{Cr} / \mathrm{Cr}^{3+}\) flows to the \(\mathrm{Pb} / \mathrm{Pb}^{2+}\) half-cell. (C) \(\mathrm{Na}^{+}\) flows to the \(\mathrm{Cr} / \mathrm{Cr}^{3+}\) half-cell, and \(\mathrm{Cl}^{-}\) flows to the \(\mathrm{Pb} / \mathrm{Pb}^{2+}\) half- cell. (D) \(\quad \mathrm{Na}^{+}\) flows to the \(\mathrm{Pb} / \mathrm{Pb}^{2}\) half- cell, and \(\mathrm{Cl}^{-}\) flows to the \(\mathrm{Cr} / \mathrm{Cr}^{3+}\) half- cell.

$$\mathrm{Br}_{2}(g)+\mathrm{I}_{2}(g) \leftrightarrow 2 \mathrm{IBr}(g)$$ At \(150^{\circ} \mathrm{C},\) the equilibrium constant, \(K_{c},\) for the reaction shown above has a value of \(300 .\) This reaction was allowed to reach equilibrium in a sealed container and the partial pressure due to IBr(g) was found to be 3 atm. Which of the following could be the partial pressures due to \(\operatorname{Br}_{2}(g)\) and \(I_{2}(g)\) in the container? \(\begin{array}{lll}{} & {\operatorname{Br}_{2}(g)} & {\mathrm{I}_{2}(g)} \\\ {\text { (A) }} & {0.1 \mathrm{atm}} & {0.3 \mathrm{atm}} \\ {\text { (B) }} & {0.3 \mathrm{atm}} & {1 \mathrm{atm}} \\ {\text { (C) }} & {1 \mathrm{atm}} & {1 \mathrm{atm}} \\ {\text { (D) }} & {1 \mathrm{atm}} & {3 \mathrm{atm}}\end{array}\)

150 \(\mathrm{mL}\) of saturated \(\mathrm{SrF}_{2}\) solution is present in a 250 \(\mathrm{mL}\) beaker at room temperature. The molar solubility of \(\mathrm{SrF}_{2}\) at 298 \(\mathrm{K}\) is \(1.0 \times 10^{-3} \mathrm{M}\) . If some of the solution evaporates overnight, which of the following will occur? (A) The mass of the solid and the concentration of the ions will stay the same. (B) The mass of the solid and the concentration of the ions will increase. (C) The mass of the solid will decrease, and the concentration of the ions will stay the same. (D) The mass of the solid will increase, and the concentration of the ions will stay the same.

Use the following information to answer questions 25-28. A voltaic cell is created using the following half-cells: \(\begin{array}{ll}{\mathrm{Cr}^{3+}+3 e \rightarrow \mathrm{Cr}(s)} & {E^{\circ}=-0.41 \mathrm{V}} \\ {\mathrm{Pb}^{2+}+2 e \rightarrow \mathrm{Pb}(s)} & {E^{\circ}=-0.12 \mathrm{V}}\end{array}\) The concentrations of the solutions in each half-cell are 1.0 M. Based on the given reduction potentials, which of the following would lead to a reaction? (A) Placing some \(\operatorname{Cr}(s)\) in a solution containing \(\mathrm{Pb}^{2+}\) ions (B) Placing some \(\mathrm{Pb}(s)\) in a solution containing \(\mathrm{Cr}^{3+}\) ions (C) Placing some \(\mathrm{Cr}(s)\) in a solution containing \(\mathrm{Cr}^{3+}\) ions (D) Placing some \(\mathrm{Pb}(s)\) in a solution containing \(\mathrm{Pb}^{2+}\) ions

$$\begin{array}{|c|c|}\hline \text { Time (Hours) } & {[\mathrm{A}] M} \\\ \hline 0 & {0.40} \\ \hline 1 & {0.20} \\ \hline 2 & {0.10} \\ \hline 3 & {0.05} \\ \hline\end{array}$$ Reactant A underwent a decomposition reaction. The concentration of A was measured periodically and recorded in the chart above. Based on the data in the chart, which of the following is the rate law for the reaction? (A) Rate \(=k[\mathrm{A}]\) (B) Rate \(=k[\mathrm{A}]^{2}\) (C) Rate \(=2 k[\mathrm{A}]\) (D) Rate \(=\frac{1}{2} k[\mathrm{A}]\)
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